Need help on understanding the solution (Linear Algebra done right)












2














In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.



It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$



It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$



For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$



Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.



I don't quite follow the logic. Can anyone help?






linear-algebra







share|cite|improve this question



























    2














    In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.



    It said:
    Note that:
    $$(a+bi)+(a-bi)=2a$$ , and
    $$(a+bi)(a-bi)=a^2+b^2$$



    It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$



    For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
    and
    $$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$



    Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.



    I don't quite follow the logic. Can anyone help?






    linear-algebra







    share|cite|improve this question

























      2












      2








      2







      In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.



      It said:
      Note that:
      $$(a+bi)+(a-bi)=2a$$ , and
      $$(a+bi)(a-bi)=a^2+b^2$$



      It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$



      For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
      and
      $$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$



      Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.



      I don't quite follow the logic. Can anyone help?






      linear-algebra







      share|cite|improve this question













      In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.



      It said:
      Note that:
      $$(a+bi)+(a-bi)=2a$$ , and
      $$(a+bi)(a-bi)=a^2+b^2$$



      It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$



      For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
      and
      $$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$



      Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.



      I don't quite follow the logic. Can anyone help?






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 1 at 8:18









      JOHN

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      676






















          4 Answers
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          If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
          $$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
          This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
          So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.






          share|cite|improve this answer





























            5














            If $z_1$ and $z_2$ are roots of



            $$(z-z_1)(z-z_2)=0$$



            $$z^2-(z_1+z_2)z+z_1z_2=0$$



            In particular, $z_1$ and $bar{z_1}$ are roots of



            $$z^2-2Re(z_1)z+|z_1|^2=0$$



            Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$



            Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$






            share|cite|improve this answer





























              2














              The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.



              Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$



              $frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$



              The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$






              share|cite|improve this answer





























                0














                If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.






                share|cite|improve this answer





















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                  4 Answers
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                  active

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                  4 Answers
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                  active

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                  active

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                  active

                  oldest

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                  3














                  If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
                  $$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
                  This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
                  So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.






                  share|cite|improve this answer


























                    3














                    If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
                    $$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
                    This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
                    So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.






                    share|cite|improve this answer
























                      3












                      3








                      3






                      If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
                      $$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
                      This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
                      So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.






                      share|cite|improve this answer












                      If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
                      $$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
                      This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
                      So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 1 at 8:47









                      Quaternion

                      1966




                      1966























                          5














                          If $z_1$ and $z_2$ are roots of



                          $$(z-z_1)(z-z_2)=0$$



                          $$z^2-(z_1+z_2)z+z_1z_2=0$$



                          In particular, $z_1$ and $bar{z_1}$ are roots of



                          $$z^2-2Re(z_1)z+|z_1|^2=0$$



                          Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$



                          Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$






                          share|cite|improve this answer


























                            5














                            If $z_1$ and $z_2$ are roots of



                            $$(z-z_1)(z-z_2)=0$$



                            $$z^2-(z_1+z_2)z+z_1z_2=0$$



                            In particular, $z_1$ and $bar{z_1}$ are roots of



                            $$z^2-2Re(z_1)z+|z_1|^2=0$$



                            Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$



                            Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$






                            share|cite|improve this answer
























                              5












                              5








                              5






                              If $z_1$ and $z_2$ are roots of



                              $$(z-z_1)(z-z_2)=0$$



                              $$z^2-(z_1+z_2)z+z_1z_2=0$$



                              In particular, $z_1$ and $bar{z_1}$ are roots of



                              $$z^2-2Re(z_1)z+|z_1|^2=0$$



                              Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$



                              Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$






                              share|cite|improve this answer












                              If $z_1$ and $z_2$ are roots of



                              $$(z-z_1)(z-z_2)=0$$



                              $$z^2-(z_1+z_2)z+z_1z_2=0$$



                              In particular, $z_1$ and $bar{z_1}$ are roots of



                              $$z^2-2Re(z_1)z+|z_1|^2=0$$



                              Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$



                              Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 1 at 8:26









                              Siong Thye Goh

                              98.4k1463116




                              98.4k1463116























                                  2














                                  The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.



                                  Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$



                                  $frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$



                                  The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$






                                  share|cite|improve this answer


























                                    2














                                    The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.



                                    Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$



                                    $frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$



                                    The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$






                                    share|cite|improve this answer
























                                      2












                                      2








                                      2






                                      The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.



                                      Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$



                                      $frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$



                                      The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$






                                      share|cite|improve this answer












                                      The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.



                                      Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$



                                      $frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$



                                      The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 1 at 9:37









                                      Shubham Johri

                                      3,433716




                                      3,433716























                                          0














                                          If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.






                                          share|cite|improve this answer


























                                            0














                                            If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.






                                            share|cite|improve this answer
























                                              0












                                              0








                                              0






                                              If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.






                                              share|cite|improve this answer












                                              If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 1 at 8:25









                                              Theoop

                                              1




                                              1






























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                                                  Алькесар

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                                                  Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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